Question Number 85456 by jagoll last updated on 22/Mar/20
$$\int\:\sqrt{\mathrm{csc}\:\mathrm{x}\:}\:\mathrm{dx}? \\ $$
Commented by jagoll last updated on 22/Mar/20
$$\mathrm{thank}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 22/Mar/20
$$\mathrm{sadly}\:\mathrm{this}\:\mathrm{is}\:\mathrm{wrong}. \\ $$$${t}=\sqrt{\mathrm{sin}\:{x}}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{\mathrm{sin}\:{x}}}{\mathrm{cos}\:{x}}{dt}=\frac{\mathrm{2}{t}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}{dt} \\ $$$$\Rightarrow\:\mathrm{2}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{4}} }}\:\mathrm{and}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this} \\ $$
Commented by MJS last updated on 22/Mar/20
![you should always test your solutions (d/dx)[2arcsin (√(sin x))]=((cos x)/( (√(sin x −sin^2 x))))≠(1/( (√(sin x))))](https://www.tinkutara.com/question/Q85464.png)
$$\mathrm{you}\:\mathrm{should}\:{always}\:\mathrm{test}\:\mathrm{your}\:\mathrm{solutions} \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{2arcsin}\:\sqrt{\mathrm{sin}\:{x}}\right]=\frac{\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{sin}\:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}}\neq\frac{\mathrm{1}}{\:\sqrt{\mathrm{sin}\:{x}}} \\ $$
Commented by jagoll last updated on 22/Mar/20
$$\mathrm{so}\:\mathrm{what}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}\:\mathrm{sir}? \\ $$
Commented by MJS last updated on 22/Mar/20
$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{it} \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{using}\:\Gamma\left({x}\right)\:\mathrm{somehow},\:\mathrm{it}\:\mathrm{had} \\ $$$$\mathrm{been}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum},\:\mathrm{maybe}\:\mathrm{a}\:\mathrm{year}\:\mathrm{ago},\:\mathrm{I}\:\mathrm{can}'\mathrm{t} \\ $$$$\mathrm{find}\:\mathrm{it} \\ $$
Commented by jagoll last updated on 22/Mar/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 22/Mar/20
![found this: ∫(dx/( (√(sin x))))= [t=x−(π/2) → dx=dt] ∫(dt/( (√(cos t))))=∫(dt/( (√(1−2sin^2 (t/2)))))= =F ((t/2)∣2) =F ((x/2)−(π/4)∣2) +C it′s an eliptic integral but I cannot explain, it′s not my solution](https://www.tinkutara.com/question/Q85483.png)
$$\mathrm{found}\:\mathrm{this}: \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}−\frac{\pi}{\mathrm{2}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$\int\frac{{dt}}{\:\sqrt{\mathrm{cos}\:{t}}}=\int\frac{{dt}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\frac{{t}}{\mathrm{2}}}}= \\ $$$$=\mathrm{F}\:\left(\frac{{t}}{\mathrm{2}}\mid\mathrm{2}\right)\:=\mathrm{F}\:\left(\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mid\mathrm{2}\right)\:+{C} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{an}\:\mathrm{eliptic}\:\mathrm{integral} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{explain},\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{my}\:\mathrm{solution} \\ $$